A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

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A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

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Answer 1 · 2017-04-11T01:38:11

The temperature at which the volume increases to $1.12 L$ $"1.12 L"$ is $6.1 \times 10^{2}$ $6.1xx10^2$ $. K$ $color(white)(.)"K"$ .

Explanation:

This is an example of Charles' law , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.

The equation to use is:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$

where $V_{1}$ $V_1$ is the initial volume, $T_{1}$ $T_1$ is the initial temperature, $V_{2}$ $V_2$ is the final volume, and $T_2$ $"T_2"$ is the final temperature.

Given/Known
$T_{1} = 362 K$ $T_1="362 K"$
$V_{1} = 0.67 L$ $V_1="0.67 L"$
$V_{2} = 1.12 L$ $V_2="1.12 L"$

Unknown: $T_{2}$ $T_2$

Solution
Rearrange the equation to isolate $T_{2}$ $T_2$ , then substitute the known values into the equation and solve.

$T_{2} = \frac{V_{2} T_{1}}{V_{1}}$ $T_2=(V_2T_1)/V_1$

$T_{2} = \frac{1.12 L \times 362 K}{0.67 L} = 6.1 \times 10^{2}$ $T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2$ $. K$ $color(white)(.) "K"$
( $605 K$ $"605 K"$ rounded to two significant figures)

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A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

1 Answer

Explanation:

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